Chapter 5.2 Polygonal Curves

Area and Orientation of Triangles

The goals of this section are:

  • to find a suitable definition for a polygon and its boundary that generalizes the definition of a triangle and a rectangle;
  • to extend the idea of orientation of a rectangular and a triangular boundary to general polygonal boundaries;
  • to discover an efficient computational method for determining the orientation of any polygonal boundary;
  • to find an explicit formula for the area of a polygon.

Triangles, as we will soon see, are the elementary building blocks of polygons and they allow us to decompose polygons in useful ways.

We follow this strategy for determining a formula for area:

  • Require that area has certain properties that agree with our physical intuition about area;
  • Use these properties to derive a formula for area;
  • Show that the formula for area actually has the desired properties.

It turns out that there will only be one formula for the area of a polygon that endows area with these properties.

We cannot assume that area is defined for any subset of the plane—This is asking for too much.

Area of a Subset of \(\mathbb{R}^2\)

For any subset \(X\) of \(\mathbb R^2\) for which area is defined, denote by \({\mathcal A}(X)\) the area of \(X\).

Area should agree with our physical intuition.

Area of a Rectangle

If \(R\) is a rectangle with side lengths \(a\) and \(b\), then \({\mathcal A}(R)\) should be defined and \[{\mathcal A}(R) = ab.\]

Now we will talk about axioms we require.

Area Axioms

Area Axioms:

For any subsets of the plane \(X\) and \(Y\) for which \({\mathcal A}(X)\) and \({\mathcal A}(Y)\) are defined, require the following:

  1. (non-negativity) \({\mathcal A}(X)\) is non-negative;
  2. (invariance) For any rigid motion of the plane \(T\), \[{\mathcal A}(T(X)) = {\mathcal A}(X);\]
  3. (closure) \({\mathcal A}(X\cup Y)\), \({\mathcal A}(X\cap Y)\), and \({\mathcal A}(X\setminus Y)\) are defined where \(X\setminus Y=\{x\in X\text{ and } x\not\in Y\}\);
  4. (additivity) If \(X\) and \(Y\) have empty intersection, then \[{\mathcal A}(X\cup Y) ={\mathcal A}(X) + {\mathcal A}(Y).\]

Use the pictures below to understand the axioms.

What follows are some immediate consequences. For example, the empty set has zero area.

Example 1

Show that the area of the empty set is \(0\).

Take \(X\) to be a square with side length 1 vertices at \((0,0)\), \((1,0)\), \((1,1)\) and \((0,1).\)

Take \(Y\) to be a square with side length 1 vertices at \((2,0)\), \((3,0)\), \((3,1)\) and \((2,1).\)

Both squares have well-defined area with \(\mathcal{A}(X)=1\) and \(\mathcal{A}(Y)=1\). Their intersection is \(X\cap Y=\emptyset\). By closure axiom, \(\mathcal{A}(\emptyset)\) is defined.

Suppose that \(X\) is a square with side length 1. Notice that \(X\cup\emptyset=X\) and \(X\) and \(\emptyset\) have empty intersection. The additivity axiom implies \[ \begin{align*} \mathcal{A}(X\cup \emptyset)&=\mathcal{A}(X)+\mathcal{A}(\emptyset)\\ \mathcal{A}(X)&=\mathcal{A}(X)+\mathcal{A}(\emptyset)\\ 0&=\mathcal{A}(\emptyset). \end{align*} \]

The example shows that area respects containment

Example 2

Show that for any subsets \(X\) and \(Y\) for which \({\mathcal A}(X)\) and \({\mathcal A}(Y)\) are defined, \[X\subseteq Y \quad \text{implies that}\quad {\mathcal A}(X) \leq {\mathcal A}(Y).\]

Take \(X\) and \(Y\) to be subsets for which \({\mathcal A}(X)\) and \({\mathcal A}(Y)\) are defined and \(X\subseteq Y\).

Take \(W=Y\setminus X\).

Notice that \(X\cap W=\emptyset\) and \(X\cup W=Y\). Thee additivity axiom implies

\[ \begin{align*} \mathcal{A}(X\cup W)&=\mathcal{A}(X)+\mathcal{A}(W)\\ \mathcal{A}(Y)&=\mathcal{A}(X)+\mathcal{A}(W)\\ \end{align*} \]

Because of the non-negativity axiom, all quantities \(\mathcal{A}(Y), \mathcal{A}(X), \mathcal{A}(W)\) are non-negative. Therefore \[\mathcal{A}(X)+\mathcal{A}(W)\geq A(X).\]

Conclude that \(\mathcal{A}(Y)\geq A(X)\) or \({\mathcal A}(X) \leq {\mathcal A}(Y).\)

The next example shows that area has the subadditvity property. It also provides a way to estimate area of a region that is the union of two sets. This is different from additivity because here we do not assume that the intersection of the two sets is empty.

Example 3

Show that for any subsets \(X\) and \(Y\) for which \({\mathcal A}(X)\) and \({\mathcal A}(Y)\) are defined, \[{\mathcal A}(X\cup Y) \leq {\mathcal A}(X) + {\mathcal A}(Y).\] This is known as the subadditivity of area.

Take \(X\) and \(Y\) to be subsets for which \({\mathcal A}(X)\) and \({\mathcal A}(Y)\) are defined.

Define \(W=Y\setminus X\). Notice \(X\cup Y=X\cup W\) and \(X\cap W=\emptyset\). The additivity axiom, implies that

\[ \begin{align*} \mathcal{A}(X\cup W)&=\mathcal{A}(X)+\mathcal{A}(W)\\ \mathcal{A}(X\cup Y)&=\mathcal{A}(X)+\mathcal{A}(W). \end{align*} \]

where \(W=Y\setminus X\), \(W\subseteq Y\). From the pervious example, \(\mathcal{A}(W)\leq \mathcal{A}(Y)\). Use that to conclude that

\[ \begin{align*} \mathcal{A}(X\cup Y)&=\mathcal{A}(X)+\mathcal{A}(W)\\ \mathcal{A}(X\cup Y)&\leq \mathcal{A}(X)+\mathcal{A}(Y) \end{align*} \]

The next example extends the additivity property to a union of more than two disjoint sets.

Example 4

Show that if \(\{X_1, X_2, \dots, X_n\}\) are all pairwise disjoint and if the area of each of the sets \(X_i\) is defined, then \[{\mathcal A}\!\left(X_1\cup\cdots\cup X_n\right) ={\mathcal A}(X_1) +\cdots+{\mathcal A}(X_n).\]

Take \(\{X_1, X_2, \dots, X_n\}\) to be pairwise disjoint. This means \(X_i\cap X_j=\emptyset\) for \(i\not=j\).

Take \(C_1=X_2\cup X_3\cup\cdots\cup X_n\). Then \(\left(X_1\cup\cdots\cup X_n\right)=X_1\cup C_1\) and \(X_1\cap C_1=\emptyset.\) The additivity axioms implies

\[ \begin{align*} \mathcal{A}\!\left(X_1\cup\cdots\cup X_n\right)&=\mathcal{A}(X_1)+\mathcal{A}(C_1)\\ &=\mathcal{A}(X_1)+\mathcal{A}(X_2\cup X_3\cup\cdots\cup X_n).\\ \end{align*} \]

Take \(C_2=X_3\cup X_4\cup\cdots\cup X_n\). Then \(\left(X_2\cup X_3\cup\cdots\cup X_n\right)=X_2\cup C_2\) and \(X_2\cap C_2=\emptyset.\) The additivity axioms implies

\[ \begin{align*} \mathcal{A}\!\left(X_1\cup\cdots\cup X_n\right)&=\mathcal{A}(X_1)+\mathcal{A}(X_2\cup X_3\cup\cdots\cup X_n)\\ &=\mathcal{A}(X_1)+\mathcal{A}(X_2 \cup C_2)\\ &=\mathcal{A}(X_1)+\mathcal{A}(X_2)+\mathcal{A}(C_2)\\ &=\mathcal{A}(X_1)+\mathcal{A}(X_2)+\mathcal{A}(X_3 \cup X_4\cup\cdots\cup X_n)\\ \end{align*} \]

Continue this process by defining \(C_j=X_{j+1}\cup\cdots\cup X_n\) so that \(\left(X_j\cup X_{j+1}\cup\cdots\cup X_n\right)=X_j\cup C_j\) and \(X_j\cap C_j=\emptyset\). The additivity axioms implies

\[{\mathcal A}\!\left(X_1\cup\cdots\cup X_n\right) ={\mathcal A}(X_1) +\cdots+{\mathcal A}(X_n).\]

This next example demonstrates the general additivity of area.

Example 5

Show that for any subsets \(X\) and \(Y\) for which \({\mathcal A}(X)\) and \({\mathcal A}(Y)\) are defined, \[{\mathcal A}(X\cup Y) = {\mathcal A}(X) + {\mathcal A}(Y)-\mathcal{A}(X\cap Y).\]

Notice that \[X=(X\setminus Y)\cup (X\cap Y)\] and \[Y=(Y\setminus X)\cup (X\cap Y),\] where \((X\setminus Y)\cap (X\cap Y)=\emptyset\) and \((Y\setminus X)\cap (X\cap Y)=\emptyset.\)

The additivity axioms implies

\[\mathcal{A}(X)=\mathcal{A}(X\setminus Y)+\mathcal{A}(X\cap Y)\]

and

\[\mathcal{A}(Y)=\mathcal{A}(Y\setminus X)+\mathcal{A}(X\cap Y).\]

Since \((X\setminus Y)\cup (Y\setminus X)\cup(X\cap Y)=(X\cup Y)\), \((X\setminus Y)\cap (Y\setminus X)=\emptyset\), \((X\setminus Y)\cap (X\cap Y)=\emptyset\), and \((Y\setminus X)\cap(X\cap Y)=\emptyset\), the additivity axioms implies

\[ \begin{align*} \mathcal{A}(X\cup Y)&=\mathcal{A}((X\setminus Y)\cup (Y\setminus X)\cup(X\cap Y))\\ &=\mathcal{A}(X\setminus Y)+\mathcal{A}(Y\setminus X)+\mathcal{A}(X\cap Y). \end{align*} \]

Therefore,

\[ \begin{align*} \mathcal{A}(X)+\mathcal{A}(Y)&=\mathcal{A}(X\setminus Y)+\mathcal{A}(X\cap Y)+\mathcal{A}(Y\setminus X)+\mathcal{A}(X\cap Y)\\ &=\mathcal{A}(X\setminus Y)+\mathcal{A}(Y\setminus X)+\mathcal{A}(X\cap Y)+\mathcal{A}(X\cap Y)\\ &=\mathcal{A}(X\cup Y)+\mathcal{A}(X\cap Y), \end{align*} \]

implies that

\[{\mathcal A}(X\cup Y) = {\mathcal A}(X) + {\mathcal A}(Y)-\mathcal{A}(X\cap Y).\]

Because a line is “one-dimensional” it should have an area of zero. That is what we show in this example.

Example 6

Show that the area of any line segment is \(0\).

Take \(L\) to be a line segment. Create two rectangles \(X\) and \(Y\) of identical side lengths so that \(X\cap Y=L\) and \(\mathcal{A}(X)=\mathcal{A}(Y)\) and \(\mathcal{A}(X\cup Y)=2\mathcal{A}(X).\)

From the previous example, \[\mathcal{A}(X\cup Y)=\mathcal{A}(X)+\mathcal{A}(Y)-\mathcal{A}(X\cap Y).\]

Use the above to conclude that

\[ \begin{align*} \mathcal{A}(X\cup Y)&=\mathcal{A}(X)+\mathcal{A}(Y)-\mathcal{A}(X\cap Y)\\ 2\mathcal{A}(X)&=2\mathcal{A}(X)-\mathcal{A}(L)\\ 0&=\mathcal{A}(L). \end{align*} \]

A point has no “width” or “height” so its area also zero.

Example 7

Show that the area of any point in the plane is \(0\).

Take \(P\) to be a point in the plane . Create two line segments \(L_1\) and \(L_2\) of identical lengths so that \(L_1\cap L_2=P\).

The general additivity property implies that \[\mathcal{A}(L_1\cup L_2)=\mathcal{A}(L_1)+\mathcal{A}(L_2)-\mathcal{A}(L_1\cap L_2).\]

Use that to conclude

\[ \begin{align*} \mathcal{A}(L_1\cup L_2)&=\mathcal{A}(L_1)+\mathcal{A}(L_2)-\mathcal{A}(L_1\cap L_2)\\ 0&=0-\mathcal{A}(P)\\ 0&=\mathcal{A}(P). \end{align*} \]

We will use the properties of area to derive a formula for the area of a triangle in terms of the coordinates of its vertices. We can do this by using transformation.

First show that if the oriented triangle \(\Delta p_1p_2p_3\) is a right triangle with leg lengths \(a\) and \(b\), then \({\mathcal A}(\Delta p_1p_2p_3)\) exists and \[{\mathcal A}(\Delta p_1p_2p_3) = \frac{ab}{2}.\]

Create rectangle \(X\) and \(Y\) with the following characteristics:

  • Rectangle \(X\) has length \(a\) and width \(b\).
  • Rectangle \(Y\) has length \(c=\sqrt{a^2+b^2}\) and width \(b\).

Arrange \(X\) and \(Y\) in the xy plane via translation and or rotation so that

  • Rectangle \(X\) has its vertices at \((0,0)\), \((a,0)\), \((a,b)\) and \((0,b)\)
  • Rectangle \(Y\) has its vertices at \((0,0)\), \((c,0)\), \((c,-b)\) and \((0,-b)\)

Rotate \(Y\) so that vertex \((c,0)\) is at \((a,b)\), which we can do by rotating about the origin by the angle \(\left(\frac{a}{c},\frac{b}{c}\right).\)

When we take \(X\cap Y\), we will get a right triangle with leg lengths \(a\) and \(b\), meaning \(X\cap Y=\Delta p_1p_2p_3\)

By the closure axiom, \({\mathcal A}(X\cap Y)={\mathcal A}(\Delta p_1p_2p_3)\) exists.

Finally, a rectangle \(R\) with side lengths \(a\) and \(b\) can be cut diagonally so that we can create two identical copies of an oriented right triangle \(\Delta p_1p_2p_3\) with leg lengths \(a\) and \(b\). Thus, \[2\mathcal{A}(\Delta p_1p_2p_3)=\mathcal{A}(R)\] or \[\mathcal{A}(\Delta p_1p_2p_3)=\frac{ab}{2}.\]

Now we talk about the area of a general triangle.

Shoelace Formula

Take \(p_1=(x(p_1),y(p_1))\), \(p_2=(x(p_2),y(p_2))\), \(p_3=(x(p_3),y(p_3)).\) The Shoelace Formula for the Area of Triangles states that if \(\Delta p_1p_2p_3\) is positively oriented, then \[\begin{align*}{\mathcal A}(\Delta p_1p_2p_3) &= \tfrac{1}{2}\left(x(p_1)y(p_2) + x(p_2)y(p_3) + x(p_3)y(p_1)\right.\\&\hspace{1.52in} \left.-y(p_1)x(p_2)-y(p_2)x(p_3)- y(p_3)x(p_1)\right).\end{align*}\]

The formula can be derived by doing the following.

Translate the triangle so that \(p_1=(x(p_1),y(p_1))\) is at the origin. Which we can do by translating by \[\langle -x(p_1),-y(p_1)\rangle .\]

So our new vertices are at \((0,0), \left(x(p_2)-x(p_1),y(p_2)-y(p_1)\right)\) and \((x(p_3)-x(p_1),y(p_3)-y(p_1))\).

Rotate the triangle so that \((x(p_2)-x(p_1),y(p_2)-y(p_1))\) is on the \(x\)-axis, which we do by rotating by the angle \(\left(\frac{x(p_2)-x(p_1)}{\|V\|},-\frac{y(p_2)-y(p_1)}{\|V\|}\right)\), where \(V=\langle x(p_2)-x(p_1),y(p_2)-y(p_1)\rangle\).

The new vertices will be

  • \((0,0)\star \left(\frac{x(p_2)-x(p_1)}{\|V\|},-\frac{y(p_2)-y(p_1)}{\|V\|}\right)=Q_1\)
  • \((x(p_2)-x(p_1),y(p_2)-y(p_1))\star \left(\frac{x(p_2)-x(p_1)}{\|V\|},-\frac{y(p_2)-y(p_1)}{\|V\|}\right)=Q_2\)
  • \((x(p_3)-x(p_1),y(p_3)-y(p_1))\star \left(\frac{x(p_2)-x(p_1)}{\|V\|},-\frac{y(p_2)-y(p_1)}{\|V\|}\right)=Q_3\)

where \((a,b)\star (c,d)=(ac-bd,ad+bc).\)

Now we can use find the area of the triangle, by identifying the base and the height. The base will equal the \(x\)-coordinate of \(Q_2\) while the height will equal the \(y\)-coordinate of \(Q_3\).

The \(x\)-coordinate of \(Q_2\) will be

\[ \begin{align*} &(x(p_2)-x(p_1))\cdot \left(\frac{x(p_2)-x(p_1)}{\|V\|}\right)-(y(p_2)-y(p_1))\cdot\left(-\frac{y(p_2)-y(p_1)}{\|V\|}\right)\\&=\frac{(x(p_2)-x(p_1))^2}{\|V\|}+\frac{(y(p_2)-y(p_1))^2}{\|V\|}\\ &=\frac{(x(p_2)-x(p_1))^2+(y(p_2)-y(p_1))^2}{\|V\|}\\ &=\frac{\|V\|^2}{\|V\|}\\ &=\|V\|. \end{align*} \]

The \(y\)-coordinate of \(Q_3\) will be

\[ \begin{align*} &(x(p_3)-x(p_1))\cdot \left(-\frac{y(p_2)-y(p_1)}{\|V\|}\right)+(y(p_3)-y(p_1))\cdot\left(\frac{x(p_2)-x(p_1)}{\|V\|}\right)\\&=\frac{(x(p_3)-x(p_1))\cdot(-y(p_2)+y(p_1))}{\|V\|}+\frac{(y(p_3)-y(p_1))\cdot (x(p_2)-x(p_1))}{\|V\|}\\ &=\frac{(x(p_3)-x(p_1))\cdot(-y(p_2)+y(p_1)+(y(p_3)-y(p_1))\cdot (x(p_2)-x(p_1))}{\|V\|}. \end{align*} \]

Therefore, the area of the triangle is

\[ \begin{align*} {\mathcal A}(\Delta p_1p_2p_3) &=\frac{1}{2}x(Q_2)\cdot y(Q_3)\\ &=\frac{1}{2}\|V\|\cdot\frac{(x(p_3)-x(p_1))\cdot(-y(p_2)+y(p_1)+(y(p_3)-y(p_1))\cdot (x(p_2)-x(p_1))}{\|V\|}\\ &=\frac{1}{2}\left((x(p_3)-x(p_1))\cdot(-y(p_2)+y(p_1))+(y(p_3)-y(p_1))\cdot (x(p_2)-x(p_1)\right)\\ &=\frac{1}{2}\left(-x(p_3)y(p_2)+x(p_3)y(p_1)+x(p_1)y(p_2)-x(p_1)y(p_1)+y(p_3)x(p_2)-y(p_3)x(p_1)-y(p_1)x(p_2)+y(p_1)x(p_1)\right)\\ &=\frac{1}{2}\left(x(p_1)y(p_2)+x(p_2)y(p_3)+x(p_3)y(p_1)-y(p_1)x(p_2)-y(p_2)x(p_3)-y(p_3)x(p_1)\right) \end{align*} \] as desired.

The formula also provides a way to measure orientation.

Oriented Triangle

Given an oriented triangle \(\Delta p_1p_2p_3\), take \(\alpha(\Delta p_1p_2p_3)\) to be the quantity \[\begin{align*}\alpha(\Delta p_1p_2p_3) &= \tfrac{1}{2}\left(x(p_1)y(p_2) + x(p_2)y(p_3) + x(p_3)y(p_1)\right.\\&\hspace{1.52in} \left.-y(p_1)x(p_2)-y(p_2)x(p_3)- y(p_3)x(p_1)\right).\end{align*}\]

This is a positive quantity if \(\Delta p_1p_2p_3\) if positively oriented and is negative if \(\Delta p_1p_2p_3\) is negatively oriented.

Practice finding the area with the next example. Pay attention to the sign of the quantity \(\alpha\).

Example 8

For these choices of \(p_1\), \(p_2\), and \(p_3\), use the shoelace formula to determine the area of the triangle \(\Delta p_1p_2p_3\) and to determine whether or not the triangle is positively or negatively oriented:

  1. \(p_1=(1,2)\), \(p_2=(5,4)\), \(p_3=(3,10)\);

  2. \(p_1=(2,-4)\), \(p_2=(-3,-1)\), \(p_3=(-1,4)\).

In this example, \(x(p_1)=1, x(p_2)=5,x(p_3)=3\) and \(y(p_1)=2, y(p_2)=4,y(p_3)=10\). Use the formula to compute:

\[\begin{align*}\alpha(\Delta p_1p_2p_3) &= \tfrac{1}{2}\left(1\cdot 4 + 5\cdot 10 + 3\cdot 2-2\cdot 5-4\cdot 3- 10\cdot 1\right)\\ &=\tfrac{1}{2}\left(4+50+6-10-12-10\right)\\ &=\tfrac{1}{2}\left(28\right)\\ &=14.\end{align*}\] Since the quantity is positive, the triangle is positively oriented and the area is 14.

In this example, \(x(p_1)=2, x(p_2)=-3,x(p_3)=-1\) and \(y(p_1)=-4, y(p_2)=-1,y(p_3)=4\). Use the formula to compute:

\[\begin{align*}\alpha(\Delta p_1p_2p_3) &= \tfrac{1}{2}\left(2\cdot (-1) + (-3)\cdot (4) + (-1)\cdot (-4)-(-4)\cdot (-3)-(-1)\cdot (-1)- 4\cdot 2\right)\\ &=\tfrac{1}{2}\left(-2-12+4-12-1-8\right)\\ &=\tfrac{1}{2}\left(-31\right)\\ &=-\frac{31}{2}.\end{align*}\] Since the quantity is negative, the triangle is negatively oriented and the area is \(\left|-\frac{31}{2}\right|=\frac{31}{2}.\)

In the next example, we use the axiom properties of area to get the formula of a parallelogram. The parallelogram can be split so that we get two triangles that share an edge. We can calculate the area of these two triangles to get the area of the entire parallelogram.

Example 9

Take \(\square\big((1,2),(5,3),(6,6),(2,5)\big)\) to be the parallelogram with the given ordered vertex set and compute \({\mathcal A}\!\left(\square\big((1,2),(5,3),(6,6),(2,5)\big)\right).\)

Based on the picture, calculate the area of the parallelogram by calculating the area of \(\Delta((1,2)(5,3)(6,6))\) and \(\Delta((6,6)(2,5)(1,2))\). Try it and get

\[\mathcal{A}(\Delta(1,2)(5,3)(6,6))=\frac{11}{2}\]

and

\[\mathcal{A}(\Delta(6,6)(2,5)(1,2))=\frac{11}{2}.\]

Thus \(\mathcal{A}(\square\big((1,2),(5,3),(6,6),(2,5)\big)=\mathcal{A}(\Delta(1,2)(5,3)(6,6))+\mathcal{A}(\Delta(6,6)(2,5)(1,2))=\frac{11}{2}+\frac{11}{2}=11.\)

Triangles are uniquely determined by the lengths of their sides. The area should be computable only from the side lengths.

Heron’s Formula

Heron’s Formula states that if \(a\), \(b\), and \(c\) are the side lengths of a triangle, then \[{\mathcal A}(P) = \sqrt{S(S-a)(S-b)(S-c)}\quad{\rm where}\quad S=\frac{a+b+c}{2}.\]

In the next example, find the area of a triangle with given side lengths. If the side lengths are not given, then you would first need to find the lengths.

Example 10

A triangle \(P\) has side lengths 5, 9, and 7. Calculate \({\mathcal A}(P)\).

Here \(a=5,b=9,c=7\), so \(S=\frac{5+9+7}{2}=\frac{21}{2}.\) Use Heron’s Formula to get \[ \begin{align*} {\mathcal A}(P)& = \sqrt{S(S-a)(S-b)(S-c)}\\ &=\sqrt{\frac{21}{2}\left(\frac{21}{2}-5\right)\left(\frac{21}{2}-9\right)\left(\frac{21}{2}-7\right)}\\ &=\sqrt{\frac{21}{2}\left(\frac{11}{2}\right)\left(\frac{3}{2}\right)\left(\frac{7}{2}\right)}\\ &=\sqrt{\frac{4851}{16}}. \end{align*} \]

Polygonal Curves and Triangulation

Now that we understand area of triangles, we can use that to find areas of more general curves that are known as polygonal curves.

Polygonal Path and Curve

A polygonal path is a simple, closed, continuous, piecewise linear function from a closed interval \([a,b]\) to the plane.

A polygonal curve is the image of a polygonal path.

Use this picture to understand the definition.

Now we present the Jordan Curve Theorem of Polygons.

Jordan Curve Theorem for Polygons

The Jordan Curve Theorem for Polygons guarantees that a polygonal curve \(\Gamma\) decomposes the plane into three regions, a region that is inside \(\Gamma\), a region that is outside \(\Gamma\), and the curve \(\Gamma\) itself.

A polygon is the union of a polygonal curve and the set of points that are inside the curve.

Use this picture to understand the theorem and definition of polygon.

Next we present the boundary of a polygon, which divides the outside and inside.

Boundary of a Polygon

For any polygon \(P\), denote by \(\partial P\) the polygonal curve that defines \(P\).

Denote by \(\partial_o P\) the set of oriented line segments that is determined by a polygonal path that parameterizes \(\partial P\).

Note that a triangle and a rectangle are both examples of polygons, and a triangular (path) curve and a rectangular (path) curve are both examples of polygonal (paths) curves.

Understand the boundary with the picture below.

In this example, we construct a polygonal path with a given epoch.

Example 11

Take \(\big((-4,2), (-2, 5), (2,4), (-3,7)\big)\) to be the ordered vertex set of a polygonal path with epoch \((0, 2, 4, 5, 9)\). Identify the given path with a piecewise linear function, \(c\). Simulate a particle whose position at time \(t\) is \(c(t)\).

Construct linear paths that go from one vertex to the another vertex by finding vectors that translate from one vertex to the other with the specified time:

  • Path from \((-4,2)\) at \(t=0\) to \((-2,5)\) at \(t=2\): \(\frac{t}{2}\langle 2,3\rangle+(-4,2)\)

  • Path from \((-2,5)\) at \(t=2\) to \((2,4)\) at \(t=4\): \(\frac{t-2}{2}\langle 4,-1\rangle+(-2,5)\)

  • Path from \((2,4)\) at \(t=4\) to \((-3,7)\) at \(t=5\): \((t-4)\langle -5,3\rangle+(2,4)\)

  • Path from \((-3,7)\) at \(t=5\) to \((-4,2)\) at \(t=9\): \(\frac{(t-5)}{4}\langle -1,-5\rangle+(-3,7)\)

A continuous, piecewise linear path on \([0,9]\) is

\[ c(t)=\begin{cases} \frac{t}{2}\langle 2,3\rangle+(-4,2)&\text{ if }0\leq t<2\\ \frac{t-2}{2}\langle 4,-1\rangle+(-2,5)&\text{ if }2\leq t<4\\ (t-4)\langle -5,3\rangle+(2,4)&\text{ if }4\leq t< 5\\ \frac{(t-5)}{4}\langle -1,-5\rangle+(-3,7)&\text{ if }5\leq t\leq 9\\ \end{cases} \]

Here is additional practice.

Example 12

Take \(\partial_oR\) to be the oriented boundary of a polygon that is given by \[\partial_oR = \left\{\overrightarrow{(2,2)(3,3)}, \overrightarrow{(3,3)(5,2)}, \overrightarrow{(5,2)(6,5)}, \overrightarrow{(6,5)(3,7)}, \overrightarrow{(3,7)(1,4)}, \overrightarrow{(1,4)(2,2)}\right\}.\] Take \((1,8,9,10,14, 20, 25)\) to be the epoch for a polygonal path \(c\) that parameterizes \(\partial R\) and respects \(\partial_oR\). Simulate a particle whose position at time \(t\) is \(c(t)\).

Attempt this example A path is given below

\[ c(t)=\begin{cases} \frac{t-1}{7}\langle 1,1\rangle+(2,2)&\text{ if }1\leq t<8\\ (t-8)\langle 2,-1\rangle+(3,3)&\text{ if }8\leq t<9\\ (t-9)\langle 1,3\rangle+(5,2)&\text{ if }9\leq t< 10\\ \frac{(t-10)}{4}\langle -3,2\rangle+(6,5)&\text{ if }10\leq t< 14\\ \frac{(t-14)}{6}\langle -2,-3\rangle+(3,7)&\text{ if }14\leq t< 20\\ \frac{(t-20)}{5}\langle 1,-2\rangle+(1,4)&\text{ if }20\leq t\leq 25\\ \end{cases} \]

One of the reasons we study polygons is because we can decompose it into finitely many triangles. This is known as a triangulation.

Triangulation

Take \(P\) to be a polygon. A triangulation \({\mathcal T}(P)\) is a set of oriented triangles with the following property:

  • Every oriented triangle in \({\mathcal T}(P)\) has the same orientation;
  • Any two triangles in \({\mathcal T}(P)\) are either disjoint, intersect at a point, or share an edge;
  • The union of all of the underlying triangles in \({\mathcal T}(P)\) is equal to \(P\).

Understand triangulation with the picture below.

Practice creating triangulations with the next example.

Example 13

Take \(\big((1,1), (5,5), (0,4), (-3,7), (-1,-2)\big)\) to be the ordered vertex set of a polygon \(P\). Sketch a triangulation for \(P\).

Here are the vertices of \(P\).

Here is one triangulation of \(P\): \[{\mathcal T}(P)=\{\Delta((1,1),(5,5)(0,4)), \Delta((-1,-2),(1,1)(0,4)), \Delta((0,4)(-3,7)(-1,-2))\}\]

\(\Delta((1,1),(5,5)(0,4))\)

\(\Delta((-1,-2),(1,1)(0,4))\)

\(\Delta((0,4)(-3,7)(-1,-2))\)

Based on the last example, it seems like it should be possible to create a triangulation. Is that true? What more can we say?

Now we present the following facts.

Take \(P\) to be a polygon.

  • Fact 1: Every polygon has a triangulation.

  • Fact 2: Any triangulation \({\mathcal T}(P)\) for a polygon \(P\) determines an orientation for \(\partial P\) and may be chosen to agree with \(\partial_oP\).

  • Fact 3: If two oriented triangles in a triangulation \(T(P)\) share an edge \(\overline{pq}\), then one of the triangles will have \(\overrightarrow{pq}\) in its oriented boundary and one will have \(\overrightarrow{qp}\) in its oriented boundary. Furthermore, only two triangles may share the same edge

With these facts in mind, practice constructing triangulations with a specific orientation with the next example.

Example 14

Take \(\big((0,-1), (-3,1), (-1,4), (3,3), (2,0)\big)\) to be the ordered vertex set of a polygon \(P\).

  1. Identify a positively oriented triangulation for \(P\) and identify all of the triangles in this triangulation.
  2. Identify a negatively oriented triangulation for \(P\) and identify all of the triangles in this triangulation.

Here are the vertices of \(P\).

  1. Here is one triangulation that is positively oriented: \({\mathcal T}(P)=\{\Delta((0,-1),(2,0)(3,3)), \Delta((3,3),(-1,4)(0,-1)), \Delta((-1,4)(-3,1)(0,-1))\}\)

  2. Here is one triangulation that is negatively oriented: \({\mathcal T}(P)=\{\Delta((0,-1)(3,3)(2,0)), \Delta((3,3)(0,-1)(-1,4), \Delta((-1,4)(0,-1)(-3,1))\}\)

The Area of a Polygon

We now begin the first steps for defining the area of a polygon. First, we define the \(\alpha\) quantity for an edge and then a triangle.

Oriented Edge and \(\alpha\)

For any oriented edge \(\overrightarrow{pq}\), define \(\alpha\!\left(\overrightarrow{pq}\right)\) by \[\alpha\!\left(\overrightarrow{pq}\right) = x(p)y(q) - y(p)x(q).\]

For any triangle \(\Delta pqr\), \[\alpha\!\left(\Delta pqr\right) = \alpha\!\left(\overrightarrow{pq}\right) + \alpha\!\left(\overrightarrow{qr}\right) + \alpha\!\left(\overrightarrow{rp}\right).\]

The quantity \(\alpha\!\left(\Delta pqr\right)\) is actually a function on the oriented boundary of \(\Delta pqr\)

Introduce the following useful notation:

Take \(S\) to be a finite set and for each \(x\) in \(S\), take \(f(s)\) to be a real number.

Denote by the symbol \[\sum_{s\in S}f(s)\] the sum of all values \(f(s)\), where \(s\) is in \(S\).

Example 15

Take \(S\) to be the set that is given by \[S = \{(1,2), (1,4), (3,5), (2,1), (9,3), (-4, 2), (1, -5)\}.\] For any ordered pair \((a,b)\), take \(f(a,b)\) to equal \(a\) and compute the sum \[\sum_{p\in S}f(p).\]

In this example, \[f(1,2)=1,f(1,4)=1,f(3,5)=3,f(2,1)=2,f(9,3)=9,f(-4,2)=-4,f(1,-5)=1.\]

Therefore,

\[ \begin{align*} \sum_{p\in S}f(p)&=f(1,2)+f(1,4)+f(3,5)+f(2,1)+f(9,3)+f(-4,2)+f(1,-5)\\ &=1+1+3+2+9-4+1\\ &=13. \end{align*} \]

Now take \(P\) to be a polygonal curve with an ordered vertex set \((p_1, \cdots, p_n)\) and take \({\mathcal T}(P)\) to be a triangulation of \(P\).

All triangles in \({\mathcal T}(P)\) have the same orientation, and so \[\left|\sum_{\Delta pqr \in {\mathcal T}(P)} \alpha\!\left(\Delta pqr\right)\right| = \sum_{\Delta pqr \in {\mathcal T}(P)} \left|\alpha\!\left(\Delta pqr\right)\right|.\]

The area of \(P\) is the sum of the areas of all of the triangles in \({\mathcal T}(P)\), therefore \[{\mathcal A}(P) = \frac{1}{2}\left|\sum_{\Delta pqr \in {\mathcal T}(P)} \alpha\!\left(\Delta pqr\right)\right|.\]

The contribution of any internal edge of a triangle in the triangulation to the above sum is 0 because each such edge appears in two different triangles and has opposite orientation in each, as seen in this picture:

Shoelace Formula

Take \(P\) to be a polygonal curve with an ordered vertex set \((p_1, \cdots, p_n)\) and take \({\mathcal T}(P)\) to be a triangulation of \(P\). The area of \(P\) is, therefore, the function of \(\partial_oP\) that is given by \[\begin{align*}{\mathcal A}(P) = \frac{1}{2}\left(\alpha\!\left(\overrightarrow{p_1p_2}\right) + \cdots + \alpha\!\left(\overrightarrow{p_{n}p_1}\right)\right).\end{align*}\]

This is the Shoelace Formula for the area of a polygon.

If \(\big((x_1, y_1), \dots, (x_n, y_n)\big)\) is an positively oriented ordered vertex set for \(P\), then \[{\mathcal A}(P) = \frac{1}{2}\left(x_1y_2 + x_2y_3 + \cdots + x_ny_1 - y_1x_2 - y_2x_3 - \cdots - y_nx_1\right)\]

The evocative name of the theorem comes from the following visualization of the formula.

Practice using the shoelace formula with the following example.

Example 16

Take \(\big((-4,1), (-1,3), (2,2), (0,5), (-3,4)\big)\) to be the ordered vertex set of a polygon \(P\). Take \(\partial_oP\) to be the oriented boundary that is determined by the given ordered vertex set. Use the shoelace formula to determine the area of \(P\) and the orientation of \(\partial_oP\).

Here \(x_1=-4,x_2=-1,x_3=2,x_4=0,x_5=-3\) and \(y_1=1,y_2=3,y_3=2,y_4=5,y_5=4\). Use the shoelace formula to get

\[ \begin{align*} \mathcal{A}(P)&=\tfrac{1}{2}\left((-4)\cdot3+(-1)\cdot 2+(2)\cdot 5+(0)\cdot 4+(-3)\cdot(1)-(1)\cdot(-1)-(3)\cdot(2)-(2)\cdot(0)-5\cdot(-3)-4\cdot(-4)\right)\\ &=\tfrac{1}{2}\left(-12-2+10+0-3+1-6-0+15+16\right)\\ &=\tfrac{19}{2} \end{align*} \]

Use the above answer to conclude the orientation is positive and the area is \(9.5.\)

If a given polygon \(P\) is the union of two polygons \(P_1\) and \(P_2\) that overlap only along a part of their boundary, then the union of the triangulations of \(P_1\) and \(P_2\) is a triangulation of \(P\) and no triangle lies both triangulations.

The area of \(P\) is determined by the sum of areas in any triangulation of \(P\) and so \[{\mathcal A}(P) = {\mathcal A}(P_1) + {\mathcal A}(P_2).\]

Since the area of any triangle is unchanged by rigid motions, a fact that can be directly verified, the same is also true of the area of any polygon.

Example 17

Take \(\big((-4,1), (-1,3), (2,2), (0,5), (-3,4)\big)\) to be the ordered vertex set of a polygon \(P\).

  1. Use the shoelace formula to determine the area of \(P\).
  2. Sketch \(P\) and find a triangulation of \(P\).
  3. Determine the area of \(P\) by calculating the area of the triangles in the triangulation that you found in b.

  1. The conclusion of the previous example was \(\mathcal{A}(P)=\tfrac{19}{2}\).

  2. Here is \(P\).

Here is an example of triangulation: \[{\mathcal T}(P)=\{\Delta((-1,3),(2,2)(0,5)), \Delta((0,5),(-3,4)(-1,3)), \Delta((-3,4)(-4,1)(-1,3))\}\]

\(\Delta((-1,3),(2,2)(0,5)\)

\(\Delta((0,5),(-3,4)(-1,3))\)

\(\Delta((-3,4)(-4,1)(-1,3))\)

  1. \[\begin{align*}\mathcal{A}(P)&=\mathcal{A}(\Delta((-1,3),(2,2)(0,5))+\mathcal{A}(\Delta((0,5),(-3,4)(-1,3)))+\mathcal{A}(\Delta((-3,4)(-4,1)(-1,3)))\\ &=\frac{7}{2}+\frac{5}{2}+\frac{7}{2}\\ &=\frac{19}{2}. \end{align*}\]

Same answer as previous example.

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